GRADE 12 MATHEMATICS PAPER 1 MEMORANDUM

Wednesday, 07 July 2021 09:33

GRADE 12 MATHEMATICS PAPER 1 MEMORANDUM - PAST PAPERS 2017 JUNE

MATHEMATICS PAPER 1
GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
JUNE 2017

NOTE

If a candidate answered a question TWICE, mark the FIRST attempt ONLY.
Consistent accuracy(CA) applies in ALL aspects of the memorandum.
If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt.
The mark for substitution is awarded for substitution into the correct formula.

QUESTION 1

1.1.1	𝑥² − 𝑥 − 30 = 0 (𝑥 + 5)(𝑥 − 6) = 0 𝑥 + 5 = 0 or/of 𝑥 − 6 = 0 𝑥 = −5 or/of 𝑥 = 6	✓✓ factors ✓𝑥-values (3)
1.1.2	3𝑥² + 𝑥 − 1 = 0 x = -(1)± √(1)²-4(3)(-1) 2(3) x= -1 ± √13 6 ∴x=0,43 or x = -0,77	✓ substitution ✓✓ 𝑥-values (3)
1.1.3		✓ factors ✓ critical values with method ✓✓ answer (accuracy) (4)
1.1.4	3x - 5√x = 2 3x-2 =5√x (3x-2)² =(5√x)²9x²- 12x + 4 =25x 9x²- 37x +4=0 (9x-1)(x-4)=0 ∴x= 1 or x=4 9 OR 3x-5√x=2 3x-5x^½-2=0 (3x^½+1)(x^½-2)=0 x^½= 1 or x=4 3 since x^½must be >0 x=4 is the only valid answer	✓ isolating 5 x ✓ squaring both sides ✓ standard form ✓ answers ✓ testing and conclusion ✓ standard form ✓✓ factors ✓ answers ✓ conclusion (5)
1.2	y-x-6=0..................(1) (x-3)² + (y-3)² = 18..(2) from (1); y=x+6 ...(3) (3) in (2); (x-3)² + (x+6-3)² = 18 x²-6x+9 +x² +6x +9 =18 2x² =0 x=0 y=0+6 =6	✓ substitution ✓ removing brackets ✓ standard form ✓ 𝑥-value ✓ 𝑦-value (5)
1.3	1 + 1 = 7 x + 1 5 x 1 = 7 x² + 1 5 x x = 2 x² + 1 5 2x² + 2 = 5x2x² - 5x + 2 = 0 (2x-1)(x-2) = 0 x=½ or x=2 If candidate after step 3 concludes x = 2, then max of (2/5)	✓ adding denominator ✓ simplification ✓ standard form ✓ factors or formula ✓ answers (5)

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QUESTION 2

2.1.1	1 ; 5 ; 12 ; 22 ; 𝑥 ; 𝑦 4 7 10 - 1st differences 3 3 - 2nd differences T₅ = 35 and T₆ = 51	✓✓ answers (2)
2.1.2		✓ answer (4)
2.1.3	3 n² - ½n = 3432 2 3 n²- ½n - 3432 = 0 2 3n²- n - 6864 = 0 (3n+143)(n-48) = 0 n = -143 or n = 48 3	✓ equation ✓ standard form ✓ factors or formula ✓ answer n = 48 (4)
2.2.1	𝑇₂ − 𝑇₁ = 𝑇₃ − 𝑇₂ 𝑚 + √2 = 3√2 − 𝑚 2𝑚 = 2√2 𝑚 = √2 OR	✓ method ✓ answer ✓ method ✓ answer (2)
2.2.2	T₅₁ = a + 50d =-√2 + 50(2√2) =99√2	✓ value of d ✓ substitution into correct formula ✓ answer (3)
2.3	Terms between 50 and 500 divisible by 7 First term = 56 and Last term = 497	✓ identification of first and last terms ✓ substitution ✓ answer (3)
2.4.1		✓ method ✓ substitution ✓ answer
2.4.2	Yes, because -1<r<1 -1< 1 <1 3	✓ Yes ✓ reason (2)
2.4.3		✓✓ substitution ✓ simplification ✓ exponential law ✓ answer (5)
2.5	= 1+1 (1+1) (1+1+1)(1+1+1+1)(1+1+1+1+1)(1+1+1+1+1+1) = 1+2+3+4+5+6 21	✓ expansion ✓ answer (2)

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QUESTION 3

3.1.1	𝑥 = 5	answer / antwoord (1)
3.1.2	(x-5)² = 4 x-5 = ± 2 x=3 or x=7 OR x² -10x +25 = 4 x² -10x +21 = 0 (x-3)(x-7) = 0 x=3 or x=7	✓ let y = 0 ✓ square root ✓ answer (3) ✓ let = 0 ✓ factors ✓ answers
3.1.3		✓ x-intercepts ✓ y-intercepts ✓ turning point ✓ shape (4)
3.1.4	Range of f : 𝑦 ∈ [−4; ∞] 𝑜𝑟/𝑜𝑓 𝑦 ≥ −4 𝑦 ∈ 𝑅	✓ answer (1)
3.1.5	Reflection about the x-axis y= -(x - 5)² + 4	✓ answer ✓ equation (2)
3.2	x² +3 = kx - 1 x² -kx + 4 =0 For g (x) to be a tangent, roots are equal. b² -4ac = 0 k² -16 = 0 k² = 16 / (k+4)(k-4) =0 k=±4 / k=-4 or k= 4	✓ equating ✓ standard form ✓ Δ = 0 ✓ substitution ✓ answers (5)

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QUESTION 4

4.1	p = 1 and q = 2	✓ value of p ✓ value of q (2)
4.2	y= a + 2 (x+1) 4 = a + 2 0+1 a=2 y = 2 + 2 (x+1)	✓ substitution of point ✓ value of a ✓ equation (3)
4.3	Point of intersection of axes of symmetry of f is (– 1 ; 2) Point of intersection of axes of symmetry of g is: x-3 = x + 1 2x = 4 x=2 y=-1 Transformation is from (-1;2)→(2;-1) ∴3 units to the right and 3 units down	✓ point of intersection ✓ equating ✓ x-value and y-value ✓ method ✓ answer (5)

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QUESTION 5

5.1	5 = a¹ + ½ 6 ∴ a = 1 3	✓ substitution ✓ answer (2)
5.2		✓ substitution ✓ answer (2)
5.3		✓ answer (1)
5.4	h(x) = 3^xh^-1 : y = log₃x or log x log 3	✓ answer h ✓ answer h^-1 (2)
5.5	Points: (-2;9½) and (0; 3 ) 2 m= y₂ - y₁ x₂ - x₁= 19 - 1 2 2 -2 - 0 = 8 -2 =-4	✓ coordinates of A . ✓ substitution ✓ answer (3)

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QUESTION 6

6.1	effective rate = 16.08%p.a	✓ formula ✓ substitution ✓ answer (3)
6.2.1	𝐴 = 𝑃(1 + 𝑖𝑛) = 75 000(1 +12% x 8) = 𝑅147 000 Monthly installment: R147000 96 months = R1531,25	✓ substitution ✓ answer ✓ answer (3)
6.2.2	A = P(1+i)ⁿ 147000 = 75000 (1+i)⁸ (1+i)⁸ = 1.96 1 + i = ⁸√196 i = 1,087757306 - 1 i = 0,08775... rate = 8,78%	✓ substitution ✓ simplification ✓ making i subject of the formula ✓ answer (4)
6.3		✓✓✓setting up equation ✓ answer (4)

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QUESTION 7

7.1	𝑓(𝑥) = −2𝑥² 𝑓(𝑥 + ℎ) = −2(𝑥 + ℎ)² = −2(𝑥² + 2𝑥ℎ + ℎ²) = −2𝑥² − 4𝑥ℎ − 2ℎ² 𝑓′(𝑥) = lim 𝑓(𝑥 + ℎ) − 𝑓(𝑥) ℎ→0 ℎ = lim −2𝑥2 − 4𝑥ℎ − 2ℎ2 − (−2𝑥2) ℎ→0 ℎ = lim −4𝑥ℎ − 2ℎ² ℎ→0 ℎ = lim ℎ(−4𝑥 − 2ℎ) ℎ→0 ℎ = lim (−4𝑥 − 2ℎ) ℎ→0 = −4𝑥	✓ −2x² − 4𝑥ℎ − 2ℎ² ✓ substitution ✓ common factor ✓ answer (4)
7.2.1	𝑦 = 6𝑥 + 4𝑥√𝑥 𝑦 = 6𝑥 + 4𝑥3 2 𝑑𝑦 = 6 + 6𝑥½ 𝑑𝑥	✓4𝑥3 2 6 ✓ 6𝑥½✓ (3)
7.2.2		1 tr² - ½ ✓ 6 ✓✓answer (3)

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QUESTION 8

8.1	y = 1(x-1)(x-4) =(x² -2x + 1)(x-4) =x³ -4x² - 2x²+ 8x + x - 4 = x³ - 6x²+ 9x -4 ∴b + -6 and c = 9 OR 1+b+c = 4= 0 b+c = 3 ...(1) 64 + 16b + 4c - 4 = 0 16b +4c = -60 ..(2) (2)-(1) x 4 16b + 4c = 12 ∴12b = -72 b=-6 -6+c=3 ∴c=9	✓ substitution ✓ expanding ✓ answer ✓ both equations ✓ subtraction ✓ answers (3)
8.2	𝑓(𝑥) = 𝑥³ − 6𝑥² + 9𝑥 − 4 𝑓′'(𝑥) = 3𝑥² − 12𝑥 + 9 = 0 (3𝑥 − 9)(𝑥 − 1) = 0 3𝑥 − 9 = 0 𝑜𝑟 𝑥 − 1 = 0 𝑥 = 3 𝑜𝑟 𝑥 = 1 𝑦 = −4 𝐵(3 ; −4)	✓ f ′(x) = 0 ✓ factors ✓ x-values ✓ coordinates of B (4)
8.3	x<1 or x>3	x<1 x>3
8.4	𝑓(𝑥) = 3x²-12x + 9 𝑓''(𝑥) = 6x - 12 = 0 x= 2 y= -2 Point of inflection: (2 ; -2) equation of line /: y = x-4 -2 = 2-4 -2= -2	✓ f'' (x) = 0 ✓ coordinates ✓ equation of line ✓ method (4)

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QUESTION 9

9.1	height of ΔAPQ = (8-y) x = 8-y (APQ III ABC) 10 8 8x = 80 - 10y 10y = -8x + 80 y=-8 x + 8 10	✓ ratios ✓ answer(2)
9.2		✓ formula ✓ substitution (2)
9.3	A(x) = 8x - 8x ² 10 A'(x) = 16 x + 8 = 0 10 x = -8 x - 10 16 x = 5cm y = 8 (5) + 8 = 4cm 10 OR to gey y A = 8(5) - 8(5)² 10 =20cm² y = 4cm	✓ A' (x) ✓ =0 ✓ length of x ✓ length of y (4)

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QUESTION 10

10.1.1	P(A or B)' = 1-P(A or B) = 0,3	✓ answer(1)
10.1.2	P(A or B) = P(A) + P(B) 0,7 = 0,4 + k k = 0,3	✓ rule ✓ answer (2)
10.1.2	P (A∪B) = P(A) + P(B) - P(A∩B) 0,7 = 0,4 + k - P(A∩B) P(A∩B) = k - 0,3 P(A∩B) = P(A) x P(B) k-0,3 = 0,4 x k 0,6k = 0,3 k = 0,5	✓ substitution in rule ✓ answer ✓ substitution in rule ✓ answer (4)
10.2
10.2.1	m = 10 / 5 and 30 / 3 / 1 24 12 240 24 8	✓ answer m ✓ answer n (2)
10.2.2	14 x 9 = 7 or 14 x x = 7 24 x+9 20 24 x+9 30 126 = 7 or 14x = 7 24x + 216 20 24x + 216 30 168 x + 1512 = 2520 420x = 168x + 1512 168x = 1008 252x = 1512 x = 6 x= 6	9 / x x+9 x+9 ✓ equation ✓ answer (3)
10.2.3	𝑃(𝐺𝑟𝑒𝑒𝑛) = 7 + 7 24 30 = 21 40 (0,525)	✓ addition ✓ answer (2)

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TOTAL: 150

Last modified on Thursday, 08 July 2021 10:01